Question: Find the zeros of the function. Enter the solutions from least to greatest. $f(x) = (x - 4)^2 - 16$ $\text{lesser }x = $
Explanation: $\begin{aligned} (x - 4)^2 - 16&= 0 \\\\ (x-4)^2&=16 \\\\ \sqrt{(x-4)^2}&=\sqrt{16} \end{aligned}$ $\begin{aligned} x-4&=\pm4 \\\\ x&=\pm4+4 \\ \phantom{(x - 4)^2 - 16}& \\ x=0&\text{ or }x=8 \end{aligned}$ In conclusion, $\begin{aligned} \text{lesser }x &= 0 \\\\ \text{greater } x &= 8 \end{aligned}$